Two disjoint random events

Question

You roll twice with four-sided die in which the numbers one and two occur with probability $\frac{1}{3}$, and the numbers three and four each with probability $\frac{1}{6}$. Let X be the number of singles and Y the number of fours that occurred after two throws.

How do I create a table of probability function $p_{x,y}(x,y)=P\left \{X=x \wedge Y=y\right \}$?

This symbols at the end of this quations are little bit confusing to me.

$P(X=1)=\frac{1}{3}$, $P(X=2)=\frac{1}{3}$, $P(X=3)=\frac{1}{6}$ and $P(X=4)=\frac{1}{6}$.

$P(X,Y)=P(x)P(Y)=\frac{1}{3}\frac{1}{6}=\frac{1}{18}$

So do I just write Table:

$x_i$ $P(X=x_i)$

1 $2*1/3$

4 $2*1/6$

Because there are two throws or?

How do I calculate $P\left \{X+Y>0\right \}$?

Do I just add them? $P\left \{X+Y>0\right \}=\frac{1}{3}+\frac{1}{6}$

Answer 1

If X is the number of ones seen in two rolls.

Each roll has a $\frac 13$ chance of seeing a $1$

$P(X = 2) = (\frac {1}{3})^2\\ P(X = 0) = (1-\frac {1}{3})^2\\ P(X = 1)= 2(\frac 13)(\frac 23)$

If Y is the number of $4's$ seen in two rolls.

Each roll has a $\frac 16$ chance of seeing a $4$

$P(Y = 2) = (\frac {1}{6})^2\\ P(Y = 0) = (1-\frac {1}{6})^2\\ P(Y = 1)= 2(\frac 16)(\frac 56)$

Regarding $P(X+Y>0)$

Seeing a 4 and seeing a 1 on any roll are mutually exclusive events. I think it is easier to look at the possibility that you roll a 4 or a 1. The probability is $\frac 12$ on any given roll.

$P(X+Y>0) = 1-\frac 14$

Covariance:

$var(X) = \sum P(X) X^2 - E[X]^2\\ 2^2(\frac 19) + 1^2(\frac 49) - (2(\frac 19) + 1(\frac 49))^2\\ \frac {4}{9}$

$var(Y) = 2^2(\frac 1{36}) + 1^2(\frac {10}{36}) - (2(\frac 1{36}) + 1(\frac {10}{36}))^2\\ \frac {10}{36}$

$var (X+Y)= 2^2(\frac 1{4}) + 1^2(\frac {1}{2}) - (2(\frac 1{4}) + 1(\frac {1}{2}))^2\\ \frac {1}{2}$

$var (X+Y) = var(X) + var(Y) + 2cov(X,Y)\\ \frac {1}{2} = \frac {4}{9} + \frac {5}{36} + 2cov(X,Y)\\ cov(X,Y) = -\frac {1}{9}$

Answer 2

Start by noting that the question defines $X$ as the random variate representing the number of ones appearing in the outcomes of two throws. So for example, $$ P(X=2) = \frac19\\ P(X=1) = \frac49 \\P(X=0) = \frac49 $$ Next you come to the point of the problem: $P(X=1 \wedge Y=1)$ (for instance) is very different from $P(X=1) \cdot P(Y=1)$ because once you know that precisely one $1$ was rolled, that somewhat decreases the odds of precisely one $4$. To compute the $X=1,Y=1$ entry in the table you would reason:

"To get one $1$ and one $4$ either the first roll must be $1$ and the second $4$ (probability $\frac13 \cdot \frac16 = \frac1{18}$ or the first must be $4$ and the second $1$ (same probability)." Since these two outcomes are mutually exclusive we can simply add the probabilities, giving $\frac19$ for that entry in the table."

The other entries are found in a similar way.

Answer 3

You cannot say $P(X,Y)=P(X)\cdot P(Y)$ as these events are not independent.

Further, you need to account for different orderings. For example,

$$P(X=1, Y=1)=2\cdot\left(\frac{1}{3}\cdot\frac{1}{6}\right)=\frac{1}{9}$$

since we can get a one and then a four or a four and then a one.

Similarly

$$P(X=0, Y=1)=2\cdot\left(\frac{1}{2}\cdot\frac{1}{6}\right)=\frac{1}{6}$$

since we must get a two or three and then a four, or a four and then a two or a three.

You should find that the joint probability mass function is

$$\begin{array}{|c|c|c|c|} \hline X/Y& 0 & 1 & 2 \\ \hline 0& \frac{1}{4}& \frac{1}{6}&\frac{1}{36}\\ \hline 1& \frac{1}{3}& \frac{1}{9}&0\\ \hline 2& \frac{1}{9}& 0&0\\ \hline \end{array}$$

As a check, the probabilities in this table do indeed sum to one, as required.